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# load libraries
library(tidyverse)
library(magrittr)
library(kableExtra)3 Wilcoxon signed rank test II
paired observations
Example 3.1 (Wilcoxon signed rank test (paired observations)) Going back to our sleep study, now we are ready to examine whether there is enough evidence to reject a null hypothesis of median of the differences between the paired observations is equal to 0. Being able to reject a null hypothesis would mean that the drug is having an effect.
The data we recorded are shown below:
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# input data
data.sleep <- data.frame(id = 1:10,
drug = c(6.1, 6.0, 8.2, 7.6, 6.5, 5.4, 6.9, 6.7, 7.4, 5.8),
placebo = c(5.2, 7.9, 3.9, 4.7, 5.3, 7.4, 4.2, 6.1, 3.8, 7.3))
print(data.sleep) id drug placebo
1 1 6.1 5.2
2 2 6.0 7.9
3 3 8.2 3.9
4 4 7.6 4.7
5 5 6.5 5.3
6 6 5.4 7.4
7 7 6.9 4.2
8 8 6.7 6.1
9 9 7.4 3.8
10 10 5.8 7.33.1 Define the null and alternative hypothesis under the study
\(H_0:\) the median difference in the population equals to zero
\(H_1:\) the median difference in the population does not equals to zero
3.2 Test statistics: calculate difference and rank it
We start by calculating the difference between hours of sleep for each study participant. We exclude any difference that is equal to 0. The rest of values we rank in ascending order, ignoring the sign. As a result the smallest difference value, here 0.6 is ranked 1.
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# calculate pair difference and rank it
df.wilcox.signed <- data.sleep %>%
mutate(diff = drug - placebo) %>%
mutate(rank = rank(abs(diff))) %>%
print() id drug placebo diff rank
1 1 6.1 5.2 0.9 2
2 2 6.0 7.9 -1.9 5
3 3 8.2 3.9 4.3 10
4 4 7.6 4.7 2.9 8
5 5 6.5 5.3 1.2 3
6 6 5.4 7.4 -2.0 6
7 7 6.9 4.2 2.7 7
8 8 6.7 6.1 0.6 1
9 9 7.4 3.8 3.6 9
10 10 5.8 7.3 -1.5 43.3 Test statistics: sum up the ranks of the negative differences and of positive differences and denote these sums by \(T_{-}\) and \(T_{+}\) respectively
We get \(T_{-} = 40\) and \(T_{+} = 15\)
Why? If there were no differences in effectiveness between the sleeping drug and the placebo then the sums \(T_{-}\) and \(T_{+}\) would be similar. If there were a difference then one sum would be much smaller and one sum would be much larger than expected.
Code
# sum up the ranks of the positive and negative differences
data.tsum <- df.wilcox.signed %>%
mutate(sign = ifelse(diff < 0, "+", "-")) %>%
group_by(sign) %>%
summarize(T = sum(rank)) %>%
print()# A tibble: 2 × 2
sign T
<chr> <dbl>
1 - 40
2 + 153.4 Test statistics: denote the smaller sum by T and interpret the P-value
The Wilcoxon signed rank test is based on assessing whether \(T\), the smaller of \(T_{-}\) and \(T_{+}\), is smaller than would be expected by chance, under the null hypothesis that the median of the paired differences is zero.
The hypothesis is that \(T\) is equal to the sum of the ranks divided by 2, so that the smaller \(T\) the more evidence there is against the null hypothesis.
Having our \(T\) value we can check what is the probability of observing the value of \(T\) under the null hypothesis, by checking the statistical table of “Critical values for the Wilcoxon matched pairs signed rank test” found online or here.
In our example, \(T=15\) and the sample size is \(n=10\), where \(n\) is the number of non-zero differences (we had none). According to the table, the 5% percentage point is at 8. Since \(T=15 > 8\) our \(P-value > 0.05\) and we do not have enough evidence to reject the null hypothesis. There is no evidence of the sleeping drug working.
3.5 In R we use wilcox.test() function adjusting paired argument:
# run Wilcoxon signed rank test for paired observations
wilcox.test(x = data.sleep$placebo,
y = data.sleep$drug,
alternative = "two.sided",
mu = 0,
paired = TRUE,
exact = F)
Wilcoxon signed rank test with continuity correction
data: data.sleep$placebo and data.sleep$drug
V = 15, p-value = 0.2213
alternative hypothesis: true location shift is not equal to 0